What is the integral of e^{3x}x ?

The integral of e^{3x}x is 1/9 (e^{3x}*3x-e^{3x})+C

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integral of e^{3x}x

\int e^{3 x} x d x

\frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + C

Solution steps

\int e^{3 x} x d x

Apply u-substitution

: \int \frac{e ^{u} u}{9} d u

= \int \frac{e ^{u} u}{9} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= \frac{1}{9} (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 x

= \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x})

Add a constant to the solution

= \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + C

n = 0 \sum \infty \frac{1}{n + 2}

\int \frac{9 - x ^{2}}{3 x} d x

sec (sin (x))

\frac{d S}{d t} = 0.9 - \frac{15 S}{1000}

2 x^{3} d y = y (2 x^{2} - y^{3}) d x

What is the integral of e^{3x}x ?
The integral of e^{3x}x is 1/9 (e^{3x}*3x-e^{3x})+C